24 Jan
2009
24 Jan
'09
7:05 p.m.
2009/1/24 Conal Elliott
Incorrect *if* the semantics distinguishes between () and _|_ (as apparently is the choice in Haskell).
In case there was any question, section 3.17.3 of the Haskell 98 Report states,
case _|_ of { K x1 ... xn -> e; _ -> e' } = _|_
where K is a data constructor of arity n
This means that \() -> () is strict, and will behave differently if
given () or _|_.
--
Dave Menendez